Square free values of polynomials on $\mathbb{Z}[x]$

Let $f(x) \in \mathbb{Z}[x]$ be a separable polynomial (i.e. with no repeated roots) of positive degree. Set B := gcd{$ f(n) : n \in \mathbb{Z}$} and let B' be the smallest divisor of B so that B/B' is square-free. For each prime p, we denote by $p^{q_p}$ the largest power of p dividing B', and by $r_f(p)$ the number of a mod $p^{2+q_p}$ for which $f(a)/B' = 0 mod p^2$. We set $ c_f =\prod_{p}$($1-\frac{r_f}{p^{2+q_p}})$ which is the conjectural density of integers n for which f(n)/B' is squarefree.

(a)For f(x) = x(x+1)(x+2)(x+3), find $B_f$ and $B'_f$ ,and show that $r_f (p) = 4$ for $p \neq 2, 3$. Hence $c_f = R \prod_{p\neq 2,3}(1-\frac{4}{p^2})$ .Find R.

(b) Now assume that B'= 1. Show that $c_f > 0$, i.e. that $r_f (p) < p^2$ for all primes p.

Attempt: Unfortunately I couldn't even find $B_f$ and $B'_f$. I don't understand how should I compute gcd{$f(n): n\in \mathbb{Z}$} as n varies.

So, I didn't attempted the problem after that.

I would like to attempt the other parts of problem myself. So, kindly give hints for these only. I will ask you other parts later if I am struck on it.

Hint for computing $B_f$: First, you know that $B_f$ divides the value $f(n)$ for any integer $n$, by the definition of $B_f$. So, start by computing the first few values, to get a sense of what $B_f$ might be. In particular, you should be able to show using this technique that $B_f$ divides the number you guess $B_f$ is.

Once you've made this guess, you need to show that it divides $f(n)$ for every integer $n$. For this, it's helpful to split up the verification by prime powers. Recall, for example, the proof that $2$ divides $n^2+n$ for every integer $n$ -- you only need to check $2$ values of $n$ here.

Here's a worked example, for the polynomial $f(x)=x(x^3+1)(x+11)$.

1. First, compute $f(-2)=126$, $f(-1)=0$, $f(0)=0$, $f(1)=24$, $f(2)=234$. The $\gcd$ of these values is $6$, so we'll guess $B_f$ is $6$.

2. To show that $6\mid n(n^3+1)(n+11)$ for every positive integer $n$, it suffices to show that $2$ divides this quantity and that $3$ divides this quantity. Using that $$a-b \mid P(a)-P(b)$$ where $a$ and $b$ are integers and $P$ is a polynomial with integer coefficients, it suffices to show that $2$ divides $f(0)$ and $f(1)$, and that $3$ divides $f(0)$, $f(1)$, and $f(2)$. We've already computed each of these values, so this is easy to check.

So that this is a relatively complete answer, I'll add hints for the other portions as well, but they'll be spoiler-tagged, so you can avoid looking at them if you wish.

Hints for the remainder of part (a), once $B$ and $B'$ are computed:

Since these primes don't divide $B$, we seek the number of $a$ modulo $p^2$ for which $a(a+1)(a+2)(a+3)$ is a multiple of $p^2$.

The only way for this to hold is if $p^2$ divides one of these factors, or if $p$ divides two of these factors. Why can the second case not happen?

One can compute $r_2(f)$ and $r_3(f)$ by brute force.

Hint for part (b):

If $B'=1$, then $r_p(f)$ is the number of $a$ modulo $p^2$ for which $p^2\mid f(a)$. If $r_p(f)\geq p^2$, then all $a$ satisfy this. What does this say about $$?